## Algebra

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Question 1 |

If x

^{3}+ y^{3}= 9 and x + y = 3, then the value of x^{4}+y^{4}is,21 | |

0 | |

17 | |

25 |

**Algebra**

**Discuss it**

Question 1 Explanation:

x^{3}+y^{3}= (x + y) × (x^{2}− xy + y^{2}) Putting given values of x^{3}+y^{3}and (x + y) 9 = 3 × ((x+y)^{2}− 3xy) = 3 × (9 − 3xy) = 27 − 9xy 9xy = 18 xy = 2 x^{4}+ y^{4}= (x^{2}+ y^{2})^{2}- 2x^{2}y^{2}= (x^{2}+ y^{2})^{2}- 2*4 [Putting value of xy] = ((x^{}+ y^{})^{2}- 2xy)^{2}- 2*4 [Putting values of (x+y) and xy] = (9 - 4)^{2}- 2*4 = 17

Question 3 |

For any real number x the maximum value of 4−6x−x

^{2}is at x=,4 | |

6 | |

-3 | |

3 |

**Algebra**

**Discuss it**

Question 3 Explanation:

Differentiate and equate to 0
6+2x =0
x=−3

Question 4 |

If 5

^{√x}+12^{√x}=13^{√x}then value of x is,2 | |

1 | |

3 | |

4 |

**Algebra**

**Discuss it**

Question 4 Explanation:

say x=1

17!=13

for, x=2 and 3 also not possible

x=4

52+122=132

169=169

17!=13

for, x=2 and 3 also not possible

x=4

52+122=132

169=169

Question 9 |

6 | |

2 | |

3 | |

0 |

**Algebra**

**Discuss it**

Question 9 Explanation:

Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac

(ab + bc + ca)/[(a+b+c)(abc)]

as, ab+bc+ca=0

= 0

(ab + bc + ca)/[(a+b+c)(abc)]

as, ab+bc+ca=0

= 0

Question 10 |

Following instructions are to be used throughout the quiz:
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read both the statements and
Give answer (a) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
Give answer (b) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
Give answer (c) if the data either in Statement I or in Statement II alone are sufficient to answer the question.
Give answer (d) if the data even in both Statements I and II together are not sufficient to answer the question.
Give answer (e) if the data in both Statements I and II together are necessary to answer the question.
If x,y are integers, then (x

^{2}+ y^{2})^{1/2}is an integer? I) x^{2}+ y^{2}is an integer II) x^{2}- 3y^{2}= 0A | |

B | |

C | |

D | |

E |

**Algebra**

**Data Sufficiency**

**Discuss it**

Question 11 |

4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number?

630 | |

63 | |

315 | |

105 |

**Algebra**

**Discuss it**

Question 11 Explanation:

Let the number be x. Then,
4/15 of 5x/7 = 4/9 of 2x/5 + 8
4/15 * 5x/7 = 4/9 * 2x/5 + 8
4x/21 = 8x/45 + 8
x/21 = 2x/45 + 1
x/21 = (2x + 45)/45
45x = 21(2x + 45)
45x = 42x + 45*21
3x = 45*21
x = 15*21 = 315.

Question 12 |

If 2x/(1 + 1/(1 + x/(1 - x))) = 1, then find the value of x.

1/3 | |

2/3 | |

1/2 | |

3/2 |

**Algebra**

**Discuss it**

Question 12 Explanation:

We have
2x/(1 + 1/(1 + x/(1-x))) = 1
2x/(1 + 1/((1-x+x)/(1-x))) = 1
2x/(1 + 1/(1/(1-x))) = 1
2x/(1 + 1-x) = 1
2x/(2-x) = 1
2x = 2-x
3x = 2
x = 2/3.

Question 13 |

Find the value of (1-1/3)(1-1/4)(1-1/5)...(1-1/100).

1/10 | |

1/50 | |

1/20 | |

1/100 |

**Algebra**

**Discuss it**

Question 13 Explanation:

We have
(1-1/3)(1-1/4)(1-1/5)...(1-1/100)
= (2/3)(3/4)(4/5)...(99/100)
= 2/100 = 1/50.

Question 14 |

A zoo has some peacocks and some horses. If the number of heads be 48 and the number of feet be 140, then the number of peacocks will be:

23 | |

25 | |

26 | |

27 |

**Algebra**

**Discuss it**

Question 14 Explanation:

Let there be x peacocks and y horses. Then,
x + y = 48 (equation 1)
and 2x + 4y = 140
x + 2y = 70
Substituting the value of x from equation 1, we get
48 -y + 2y = 70
y = 22
Then, x = 48 - 22 = 26.

Question 15 |

A total of 324 coins of 20 paise and 25 paise make a sum of Rs 71. The number of 25-paise coins is:

104 | |

120 | |

128 | |

124 |

**Algebra**

**Discuss it**

Question 15 Explanation:

Let the number of 20-paise coins be x and the number of 25-paise coins be y. Then,
x + y = 324 (equation 1)
And,
x/5 + y/4 = 71
4x + 5y = 71*20 (Multiplying by 20 on both sides)
Substituting the value of x from equation 1, we get
4(324-y) + 5y = 71*20
1296 - 4y + 5y = 1420
y = 124.

There are 16 questions to complete.